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\title{\heiti\zihao{2} 习题15.7}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{求抛物线 $y=x^{2}$ 与直线 $x-y-2=0$ 之间的最小距离.}
\textbf{解}$1^{\circ}$\quad
直线斜率$k=1$.所以对$y$求导得$y'=2x=1$,解得$(x_0,y_0)=(\dfrac{1}{2},\dfrac{1}{4})$处斜率为$1$.从而
$$
	d=\dfrac{\left|\dfrac{1}{2}-\dfrac{1}{4}-2\right|}{\sqrt{1+1}}=\dfrac{7\sqrt{2}}{8}
$$

\textbf{解}$2^{\circ}$\quad
设该点为$(x_0,x_0^2)$,则距离
$$
	d=\dfrac{\left|x_0-x_0^2-2\right|}{\sqrt{1+1}}
$$
在$x_0=\dfrac{1}{2}$时取最小值$\dfrac{7\sqrt{2}}{8}$.

\section{将正数 $12$ 分成三个正数 $x, y, z$ 之和,使得 $u=x^{3} y^{2} z$ 为最大.}
\textbf{解}\quad
记$f(x,y,z)=x^3y^2z-\lambda(x+y+z-12)$.对$x,y,z$求偏导得:
$$
	\begin{aligned}
		 & \dfrac{\partial f}{\partial x}=3x^2y^2z-\lambda=0 \\
		 & \dfrac{\partial f}{\partial y}=2x^3yz-\lambda=0   \\
		 & \dfrac{\partial f}{\partial z}=x^3y^2-\lambda=0   \\
		 & x+y+z-12=0
	\end{aligned}
$$
由于此问题要求$x,y,z>0$,所以其必有最大值,所以该极值点为极大值点,从而$u_{max} = 6912$.

\section{求函数 $f(x, y, z)=x^{2}+y^{2}+z^{2}$ 在条件 $a x+b y+c z=1$ 下最小值.}
\textbf{解}$1^{\circ}$\quad
设$g(x,y,z)=x^2+y^2+z^2-\lambda(ax+by+cz-1)$.对$x,y,z$求偏导得
$$
	\begin{aligned}
		 & \dfrac{\partial g}{\partial x}=2x-\lambda a=0 \\
		 & \dfrac{\partial g}{\partial y}=2y-\lambda b=0 \\
		 & \dfrac{\partial g}{\partial z}=2z-\lambda c=0 \\
		 & ax+by+cz-1=0
	\end{aligned}
$$
解得$(x,y,z)=(\dfrac{\lambda a}{2},\dfrac{\lambda b}{2},\dfrac{\lambda c}{2})$,从而$\lambda=\dfrac{2}{a^2+b^2+c^2}$.故
$$
	(x,y,z)=\left(\dfrac{a}{a^2+b^2+c^2},\dfrac{b}{a^2+b^2+c^2},\dfrac{c}{a^2+b^2+c^2}\right)
$$
由于该问题一定有最小值而不存在最大值,从而该点处为极小值,从而为最小值点.最小值为
$$
	\dfrac{a^2+b^2+c^2}{(a^2+b^2+c^2)^2}=\dfrac{1}{a^2+b^2+c^2}$$

\textbf{解}$2^{\circ}$\quad
求在面上点$(x_0,y_0,z_0)$到原点最小距离平方为垂线长度平方.
$$
	d^2=\dfrac{1}{(\sqrt{a^2+b^2+c^2})^2}=\dfrac{1}{a^2+b^2+c^2}
$$

\section{求球面 $x^{2}+y^{2}+z^{2}=4$ 上与点 $(3,1,-1)$ 距离最近和最远的点.}
\textbf{解}$1^{\circ}$\quad
半径为$2$.$(3,1,-1)$与原点距离为$\sqrt{11}$,在球外部.所以具其最近的点为$2\cdot \dfrac{(3,1,-1)}{|(3,1,-1)|}$,最远的点为$-2\cdot \dfrac{(3,1,-1)}{|(3,1,-1)|}$.

\textbf{解}$2^{\circ}$\quad
设该点为$(x_0,y_0,z_0)$.令$f(x,y,z)=(x-3)^2+(y-1)^2+(z+1)^2-\lambda(x^2+y^2+z^2-4)$.对$x,y,z$求偏导得:
$$
	\begin{aligned}
		 & \dfrac{\partial f}{\partial x} =2(x-3)-2\lambda x=0 \\
		 & \dfrac{\partial f}{\partial y} =2(y-1)-2\lambda y=0 \\
		 & \dfrac{\partial f}{\partial z} =2(z+1)-2\lambda z=0
	\end{aligned}
$$
解得$(x,y,z)=(\dfrac{3}{1-\lambda},\dfrac{1}{1-\lambda},\dfrac{-1}{1-\lambda})$.又因为$x^2+y^2+z^2=4$,所以$(x,y,z)=\pm(\dfrac{6}{\sqrt{11}},\dfrac{2}{\sqrt{11}},\dfrac{-2}{\sqrt{11}})$.显然二者为极大值点与极小值点,经验证可发现最远点为$$
	(-\dfrac{6}{\sqrt{11}},-\dfrac{2}{\sqrt{11}},\dfrac{2}{\sqrt{11}})
$$
最近点为
$$
	(\dfrac{6}{\sqrt{11}},\dfrac{2}{\sqrt{11}},\dfrac{-2}{\sqrt{11}})
$$


\section{求函数 $f(x, y, z)=x y z$ 在条件 $x^{2}+y^{2}+z^{2}=1, x+y+z=0$ 下的极值.}
\textbf{解}\quad
记$g(x,y,z)=xyz-\lambda(x^2+y^2+z^2-1)+\mu(x+y+z)$.对$x,y,z$求偏导得:
$$
	\begin{aligned}
		 & \dfrac{\partial g}{\partial x}=yz+2\lambda x+\mu=0 \\
		 & \dfrac{\partial g}{\partial y}=xz+2\lambda y+\mu=0 \\
		 & \dfrac{\partial g}{\partial z}=xy+2\lambda z+\mu=0 \\
		 & x^2+y^2+z^2-1=0                                    \\
		 & x+y+z=0
	\end{aligned}
$$
解得轮换对称的一组解:$cyc \quad (x,y,z)=(\dfrac{-1}{\sqrt{6}},\dfrac{-1}{\sqrt{6}},\dfrac{2}{\sqrt{6}}),(x,y,z)=(\dfrac{1}{\sqrt{6}},\dfrac{1}{\sqrt{6}},\dfrac{-2}{\sqrt{6}})$,其分别为极大值点和极小值点,从而极大值为$\dfrac{\sqrt{6}}{18}$,极小值为$\dfrac{-\sqrt{6}}{18}$.


\section{求容积为 $V$ 的长方体形开口水箱的最小表面积.}
\textbf{解}\quad
设长宽高为$x,y,z$从而$xyz=V$,表面积为$2\cdot(xy+2yz+2zx)$.记$f(x,y,z)=2xy+4yz+4zx-\lambda(xyz-V)$.对$x,y,z$求偏导得
$$
	\begin{aligned}
		 & \dfrac{\partial f}{\partial x}=2y+4z-\lambda yz=0 \\
		 & \dfrac{\partial f}{\partial y}=2x+4z-\lambda xz=0 \\
		 & \dfrac{\partial f}{\partial z}=4y+4z-\lambda xy=0 \\
		 & xyz=V
	\end{aligned}
$$
解得$x=y=\sqrt{2}\sqrt[3]{\dfrac{V}{2}},z=\sqrt[3]{\dfrac{V}{2}}$.所以最小值为$(8\sqrt{2}+4)\cdot (\dfrac{V}{2})^{2/3}$.\par


\section{求函数 $f(x, y, z)=x y z$ 在条件$$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{r} \quad(x>0, y>0, z>0, r>0)$$下的极小值,并证明不等式 $3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^{-1} \leqslant \sqrt[3]{a b c}$, 其中 $a, b, c$ 为任意正常数.}
\subsection{}
\textbf{解}$1^{\circ}$\quad
$$
	\begin{aligned}
		\dfrac{1}{r}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} & \geqslant \dfrac{3}{\sqrt[3]{xyz}}
	\end{aligned}
$$
所以$\sqrt[3]{xyz}\geqslant 3r$,从而$xyz\geqslant 27r^3$.\par
\textbf{解}$2^{\circ}$\quad
令$f(x,y,z)=xyz-\lambda\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{r}\right)$.对$x,y,z$分别求偏导得:
$$
	\begin{aligned}
		 & \dfrac{\partial f}{\partial x}=yz+\dfrac{\lambda}{x^2}=0 \\
		 & \dfrac{\partial f}{\partial y}=xz+\dfrac{\lambda}{y^2}=0 \\
		 & \dfrac{\partial f}{\partial z}=xy+\dfrac{\lambda}{z^2}=0 \\
		 & \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{r}
	\end{aligned}
$$
解得$x=y=z=3r$.经验证,其为极小值点.从而极小值为$27r^3$.
\subsection{}
\begin{proof}
	$$
		\begin{aligned}
			\dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} & \leqslant \dfrac{1}{\dfrac{1}{\sqrt[3]{abc}}}=\sqrt[3]{abc}
		\end{aligned}
	$$
\end{proof}
\begin{proof}
	两边齐次,不妨令$abc=1$.则记$p,q,r$分别为$\ln a,\ln b,\ln c,p+q+r=0$.只要证明
    $$
    \dfrac{3}{\dfrac{1}{e^p}+\dfrac{1}{e^q}+\dfrac{1}{e^r}}\leqslant 1
    $$
    由Jensen不等式显然可知不等式成立.
\end{proof}
\begin{proof}
	由7.1,将令$x=a,y=b,z=c$带入,则$r=\dfrac{1}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}$,带入不等式即得
	$$
		\dfrac{3}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}} \leqslant \sqrt[3]{a b c}
	$$
\end{proof}
\end{document}